If $T:\mathbb{R}_3→\mathbb{R}_7$ is a linear transformation, then is the set $\operatorname{ker}(T)$ a subset of the codomain? 
If $T:\mathbb{R}_3→\mathbb{R}_7$ is a linear transformation, then is the set $\operatorname{ker}(T)$ a subset of the codomain?

My answer is 'yes', because $\operatorname{ker}(T) \subset \mathbb{R}_3 \subset \mathbb{R}_7$; however, the latter half of that statement was not marked correct.
 A: No, by definition,
$$\mathbb{R}^n := \{(x_1, \ldots, x_n) : x_1, \ldots, x_n \in \mathbb{R}\}$$
so $\mathbb{R}^3$ consists of ordered triples but $\mathbb{R}^7$ consists of ordered septuples. No triple is a septuple, so $\mathbb{R}^3 \not\subseteq \mathbb{R}^7$.
On the other hand, for $m < n$ we can identify $\mathbb{R}^m$ with a subset of $\mathbb{R}^n$, for example by naively taking an $m$-tuple $(x_1, \ldots, x_m)$ identifying it with the $n$-tuple $(x_1, \ldots, x_m, 0, \ldots, 0)$ produced by appending $n - m$ zeroes. In this case we can view $\mathbb{R}^m$ as a subset of $\mathbb{R}^n$, but we need to say we're doing so.
This clarification is mandatory because (for $m > 0$) there are many $m$-planes (through the origin) in $\mathbb{R}^n$, that is, infinitely many subspaces of $\mathbb{R}^n$ isomorphic to $\mathbb{R}^m$. (In fact, there are infinitely many ways to identify $\mathbb{R}^m$ with a given such $m$-plane, so there is an enormous amount of choice, hence we must specify which we're making.)
